What is Y in the following reaction?C_2H_5I + | vedclass

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(B) Step $1$: The first reaction is the Williamson ether synthesis.$C_2H_5I + NaOC_2H_5 \longrightarrow C_2H_5OC_2H_5 (X) + NaI$Here,$X$ is diethyl et

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$C_2H_5I$

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(B) Step $1$: The first reaction is the Williamson ether synthesis.$C_2H_5I + NaOC_2H_5 \longrightarrow C_2H_5OC_2H_5 (X) + NaI$Here,$X$ is diethyl ether $(C_2H_5OC_2H_5)$.Step $2$: The second reaction is the cleavage of ether by excess hydroiodic acid $(HI)$.$C_2H_5OC_2H_5 + 2HI \xrightarrow{\Delta} 2C_2H_5I (Y) + H_2O$Comparing this with the given reaction $X + 2HI \xrightarrow{\Delta} 2Y + H_2O$,we find that $Y$ is $C_2H_5I$.

What is $Y$ in the following reaction?
$C_2H_5I + NaOC_2H_5 \longrightarrow X + NaI$
$X + 2HI \xrightarrow{\Delta} 2Y + H_2O$

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What is $Y$ in the following reaction?$C_2H_5I + NaOC_2H_5 \longrightarrow X + NaI$$X + 2HI \xrightarrow{\Delta} 2Y + H_2O$